A block is released from rest, determine its velocity

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The discussion centers on solving a physics problem involving a block released from rest and determining its velocity upon impact with the floor using the work-energy method. Participants express confusion about integrating acceleration when it is not constant and the implications of constants of integration in their calculations. Clarifications are provided regarding the derivation of the work-energy principle and how constants can be combined and determined by initial conditions. The importance of understanding calculus in this context is also emphasized. Overall, the thread highlights the application of work-energy principles in solving dynamics problems.
Alexanddros81
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Homework Statement


14.10 Solve Prob. 12.47 by the work-energy method

12.47 When the 1.8-kg block is in the position shown, the attached spring is
undeformed. If the block is released from rest in this position, determine its velocity
when it hits the floor

Fig P12_47.jpg


Homework Equations

The Attempt at a Solution


Here is my solution to both 12.47 and 14.10.
I get in 14.10 the correct answer but I am confused with 12.47. Is it correct?
Pytels_Dynamics109.jpg
Pytels_Dynamics108.jpg
 

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When you wrote
upload_2017-12-18_13-7-12.png

you assumed ##\int a dx = a \Delta x##. Is this true if the acceleration is not constant?

What is the main topic of chapter 12?
 

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Pytels_Dynamics111.jpg


Hi! After reading a sample problem of the book I came up with the above solution.
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
 

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Alexanddros81 said:
View attachment 217021
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
Each side of the equation would have its own constant of integration. You can combine the two constants into one constant. Then, this constant is determined by the initial conditions as you have done.

Your work here is essentially the derivation of the work-energy principle for this particular problem.
 
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Ok thanks for that.
I need to revise my calculus.

Alexandros
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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